Question

A particle is executing $SHM$ on a straight line with zero initial phase. It crosses the point from where the time is considered for the motion at successive intervals $t$ and $2t$ with a speed $v$. Find the amplitude of the motion.

• A. $vt/2\pi$
• B. $vt$
• C. $vt/\pi$
• D. $2vt$

Answer 1

The correct option is C $vt/\pi$



Given,
Initial phase, $\Phi =0$
Therefore, the time will calculate from origin.
$T=(t_{0A}+t_{A0})+(t_{0A}+t_{B0})$
$=t+t=2t=\frac{2\pi }{\omega }$
From here,
$\omega =\frac{\pi }{t}$
As the particle starts from origin,
So, equation of displacement, $x-a\sin \omega t$
$v=a\omega \cos \omega t=a\omega =-a\pi /t$
Amplitude $=|a|=\frac{vt}{\pi }$
Final answer: (c)